# Distance function from x to the Cantor set

1. Sep 18, 2008

### Dragonfall

Does the said function satisfy:

(1)continuity
(2)never constant
(3)has uncountably many zeroes

1 and 3 is trivial, but I'm not sure about 2.

2. Sep 22, 2008

### Dragonfall

Hello hello

3. Sep 22, 2008

### CRGreathouse

I'm not sure how you'd even define the distance, given that every point in (0, 1) has a point in the Cantor set within epsilon for any epsilon > 0.

4. Sep 22, 2008

### Moo Of Doom

Not true. The Cantor set is not dense in (0, 1). For example the point 1/2 is at least 1/6 away from any point in the Cantor (middle-thirds) set. In fact, dist({1/2}, CantorSet} = 1/6.

5. Sep 23, 2008

### CRGreathouse

Ah... clearly I was thinking of something else. That'll teach me to post late at night!

6. Sep 23, 2008

### Doodle Bob

If x=1/2, then the distance from x to the Cantor middle-third set would be 1/6. If x=0, then the distance would be 0. Hence "not constant".

I find the the use of the word "never" strange since it sounds to be like asserting otherwise the function would be constant on, say, the Tuesdays after a new moon, but not constant all other days.

Possibly what you mean is that there are no open sets on which the function is constant.

7. Sep 24, 2008

### Dragonfall

By "never" I mean that there is no interval on which it is constant.

This is was a problem I thought up. My intuition was that since if a function is "continuous", and "never constant", each time you hit a zero you must "wave" up and down in order to hit a zero again. So this will make the number of zeros "countable". But the distance function from x to the cantor set seems to be "continuous and never constant" but has uncountably many zeros.